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Main content
Current time:0:00Total duration:4:24
AP.CALC:
FUN‑7 (EU)
,
FUN‑7.D (LO)
,
FUN‑7.D.1 (EK)
,
FUN‑7.D.2 (EK)

Video transcript

what we're going to do in this video is get some practice finding general solutions to separable differential equations so let's say that I had the differential equation dy DX the derivative of Y with respect to X is equal to e to the X over Y see if you can find the general solution to this differential equation I'm giving you a huge hint it is a separable differential equation alright so when we're dealing with a separable differential equation what we want to do is get the Y's and the D Y's on one side and then the X is and the DX is on the other side and we really treat these differentials kind of like variables which is a little hand wavy with the mathematics but that's what we will do so let's see if we multiply both sides times y so we're going to multiply both sides times Y what are we going to get we're going to get Y times the derivative of Y with respect to X is equal to e to the X and now we can multiply both sides by the differential DX multiply both of them by DX those cancel out and we are left with Y times dy is equal to e to the X DX and now we can take the integral of both sides so let us do that so what is the integral of Y dy well here we would just use the reverse power rule we would increment the exponent so it's Y to the first button so now when we take the antiderivative it will be Y squared and then we divide by that incremented exponent is equal to well the exciting thing about ezx is its antiderivative is and its derivative is 2 e to the X is equal to e to the X plus is equal to e to the X plus C and so we can leave it like this if we like in fact this right over here this isn't an explicit function Y here isn't an explicit function of X you could actually say Y is equal to the plus or minus square root of 2 times all of this business but this would be a pretty general relationship which would satisfy this separable differential equation let's do another example so let's say that we have the derivative of Y with respect to X is equal to it's equal to y squared times sine of X pause the video and see if you can find the general solution here so once again we want to separate our wise and our X's so let's see we can multiply both sides times y to the negative 2 power y to the negative 2 y to the negative 2 these become 1 and then we can also multiply both sides times DX so if we multiply DX here those cancel out and then we multiply to DX here and so we're left with Y to the negative 2 power times dy is equal to sine of X DX and now we just can integrate both sides now what is the antiderivative of Y to the negative 2 well once again we use the reverse power rule we increment the exponent so it's going to be Y to the negative 1 and then we divide by that newly incremented exponent so we've divided by negative 1 well that would just make this thing negative that is going to be equal to so what's the antiderivative of sine of X well it is you might recognize it if I put a negative there and a negative there the antiderivative of negative sine of X well that's cosine of X so this whole thing is going to be negative cosine of X or another way to write this I could multiply both sides times a negative 1 and so these would both become positive and so I could write 1 over Y is equal to cosine of X and actually let me write it this way plus C I want to forget my plus C's plus C or I can take the reciprocal of both sides if I want to solve explicitly for Y I could get Y is equal to 1 over cosine of X plus C as our general solution and we're done that was strangely fun
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